For the answer to the question above,
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3
<span>1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 </span>
<span>1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 </span>
<span>1 + 9x + 36x^2 + 84x^3 </span>
<span>In my experience, up to the x^3 is often adequate to approximate a route. </span>
<span>(1+x) = 0.98 </span>
<span>x = 0.98 - 1 = -0.02 </span>
<span>Substituting: </span>
<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 </span>
<span>approximation = 0.834 </span>
<span>Checking the real value in your calculator: </span>
<span>(0.98)^9 = 0.834 </span>
<span>So you have approximated correctly. </span>
<span>If you want to know how accurate your approximation is, write out the result of each in full: </span>
<span>1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 </span>
<span> (0.98)^9 = 0.8337477621 </span>
<span>So it is correct to 4</span>
Answer:
The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>
Step-by-step explanation:
The given data are;
Bakery A
<em> </em>= 1,880 cal
s₁ = 148 cal
n₁ = 10
Bakery B
<em> </em>= 1,711 cal
s₂ = 192 cal
n₂ = 10
df = n₁ + n₂ - 2
∴ df = 10 + 18 - 2 = 26
From the t-table, we have, for two tails, 
= 1.706
≈ 178
Therefore, we get;
Which gives;
Therefore, by rounding to the nearest integer, we have;
The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289
6n +5=12. The answer is a
Answer:
The ascending order is 
Step-by-step explanation:
First of all find the exact value of each number.
The fourth number is 435.
Using the above calculation we can easily arrange these numbers form least to greatest.
The ascending order is
It can be written as
.
